I'm trying to solve for the function this limit approaches:
$$\lim_{x\to \infty} \frac{x^2}{x-2}$$
From graphing the function online I know it approaches $y=x+2$, but I am just unsure how to go about proving it. I can understand why it approaches $y=x$ but I'm unsure as to where the $+2$ comes from.
Any help would be greatly appreciated
Let us prove your statement by simply showing $$\frac{x^2}{x-2}=x+\frac{4}{x-2}+2$$
So $$\frac{x^2}{x-2}=\frac{x^2-4+4}{x-2}=\frac{x^2-4}{x-2}+\frac{4}{x-2}=\frac{(x-2)(x+2)}{x-2}+\frac{4}{x-2}=x+2+\frac{4}{x-2}$$
Now take the limit as $x\to\infty$ of both sides: $$\lim_{x\to\infty}\frac{x^2}{x-2}=\lim_{x\to\infty}\left(x+2+\frac{4}{x-2}\right)=\lim_{x\to\infty}\left(x+2\right)+\lim_{x\to\infty}\frac{4}{x-2}=\lim_{x\to\infty}\left(x+2\right)$$